
Re: 136 theorems on 29 pages
Posted:
Sep 28, 2012 4:44 PM


Waldek Hebisch schrieb: > > clicliclic@freenet.de wrote: > > > > Waldek Hebisch schrieb: > > > > > > clicliclic@freenet.de wrote: > > > > > > > > In my view, the simple, trivial, answer as to why such formulae > > > > exist is that the "method of undetermined coefficients" works: In > > > > other words, one may insert polynomial factors and ramp up their > > > > degree until the linear system of equations that results after > > > > differentiation becomes soluble. > > > > > > Do you have a proof that this works? It seem that if you incresase > > > degree only on one side, then you end up with multiterm reccurences. > > > If you increase degree on both sides you have more equations. > > > > You are overlooking that there are three sides to the equation: the > > given integral, the transformed integral, and the integrated term. Extra > > "numerator" polynomials are inserted into each: the coefficients of that > > in the given integrand are assumed to be given, those of the other two > > assumed to be unknown. For a recurrence formula, the polynomials in the > > two integrands should have the same degree, and the highest admissible > > degree of that in the integrated term will obviously correlate with > > that. When the degrees are ramped up simultaneously, the numbers of > > unknowns rises twice as fast as the number of knowns, and a solution > > must eventually become feasible. This (fairly obvious) argument could no > > doubt be formalized into a mathematical proof, but I doubt the author > > would be interested in this. > > What is obvoius is that you do not have a proof  you deal with > inhomogeneous system and merely counting number of equations > is not enough to proof consistency of such system. > > And there is quite trivial reason to be suspect of this argument. > Namely, it does not use assumptions about roots.
I am not particularly interested in a formal proof of this (as opposed to being interested computeralgebra systems that produce continuous antiderivatives where possible, and can handle up to and including elliptic integrands), and the author of the paper doesn't look very interested either. I agree, of course, that not every linear system of n equations in n unknowns has solutions. However, for unrestricted polynomial powers, the roots (and exponents) are unrestricted, and the formulae in the paper fail precisely in the case of special relations among the polynomial coefficients (and exponents). This means that the equations can become dependent, which is no surprise.
Anybody who considers the existence of analogous formulae for higher degrees in doubt, should certainly try to provide a more formal proof. He may find the following basic observations helpful: (i) The existence of singleexponent singlestep recurrence formulae implies the existence of such formulae for any combination of exponent steps (i.e. raising or lowering of more than one exponent in single or multiple steps). This follows by iterating the recurrence formulae. (ii) Only relations involving linear polynomials need be considered, since the "denominator" polynomials can always be decomposed into linear factors. The piecewise constant factors caused by branchcut crossings just drop out of the relations. (iii) This decomposition automatically takes care of all degenerate cases, which are turned into lowerorder nondegenerate cases.  What remains to be analyzed, then, are the singlestepup and singlestepdown cases for n linear factors. The minimum degree of the "numerator" polynomials in the integrands will rise linearly with the number of linear factors (it turns out to be the number of factors minus two).
This looks like it could make a nice student exercise. If, on the other hand, somebody just needs more recurrence relations of this type, or wants to code a general procedure for friCAS perhaps, he should just apply the method of undetermined coefficients (and report any failures please!).
> > > > > > > > > [...] Also, there is very little practical interest in going beyond > > > > elliptic integrals (or beyond Lauricella FD). > > > > > > Hmm, that is your opinion. Note however, that when integrating > > > functions of form R*sqrt(P) where R is a rational function and P is > > > of degree 3, the rules apply only if R is of degree 1. This seems to > > > be quite serious limitation. Of course you can split R into partial > > > fractions, but to get denominator which is power of linear factor may > > > require extra algebraic extension (in particular may require going to > > > complex numbers). > > > > Let's say, this summarizes my personal experience in science and > > engineering. I agree with the author that the x^1 and x^3 terms in an > > elliptic radicand are best annihilated right away to simplify the > > problem. If the result involves elliptic integrals of the third kind, > > Pi(x,n,k), the rational cofactor has to be split into partial fractions > > anyway in order to extract the parameters n, which may therefore come in > > complex conjugate pairs, etc. > > There are lucky cases where reduction process completely solves > the integral. In other cases, you may need algebraic extensions > for logarithmic and Pi terms, but algebraic part can be still > computed without extra extensions. >
Agreed. But cases where this is absolutely crucial appear to be rare when problems relating to the physical world are addressed by integration.
> > > > One of the pseudoelliptic examples (here taken from Goursat's Cours > > d'analyse) is given on page 50 of the booklet mentioned. > > AFAICS this example is wrong  FriCAS result indicates that > there is no elementary integral in this case and I see no > reasons why it should have one. OTOH, if one modifies > the example to have > > f/sqrt((1  x^2)*(1  k^4*x^2)) > > under integral, then FriCAS can handle it for particular fs. >
Thanks for trying. My GradshteynRyzhik has three such integrands and adds more detail (without specifying the source). In particular, the substitution z*x = SQRT(x*(1x)*(1k^2*x)) is said to turn the integrand into a rational function. I will look into this claim. The other two examples are the same integrand f(x)/sqrt(x*(1x)*(1k^2*x)) with f(x) + f((1k^2*x)/(k^2*(1x))) = 0 and with f(x) + f((1x)/(1k^2*x)) = 0.
> > > > > > Well, there is more conditions beyond confluent roots, so just > > > excluding them is not enough  the user still have to check for > > > division by zero. However, the point of my remark was that repeated > > > blurb about confluent roots is more distraction than help for the > > > reader. Of course, the author is free to ignore my opinion... > > > > I expect your opinion will be appreciated by the author. As far as I can > > see, excluding additional confluences at any level of degeneracy is a > > sufficient condition (though not a necessary condition) to prevent zero > > division: Some confluences play no role (e.g. 4*d*f = e^2 doesn't affect > > 8.1.1, 8.1.2, 8.1.3 in the paper at all) whereas others play a selective > > role (e.g. 4*a*c = b^2 affects 8.1.1 and 8.1.3 in the paper but doesn't > > affect 8.1.2). While simply excluding confluent roots then seems to be > > enough indeed, it also does no harm since those integrands are dealt > > with separately anyway. > > AFAICS there are three kinds of conditions: > > 1) Nonzero leading terms (assumed to hold globally) > 2) confluent roots > 3) exceptional values of exponents, for example 3.3.2 contains > factor (n + 1)*(2*n + 1), so 1 and 1/2 must be excluded. > > In the intended range, that is increasing exponents toward > 1 or decreasing them toward 0 all exponents are good. But > outside of that range there are exceptional values. >
I see. It looked like you were concerned about some problem with the discriminants/resultants, which can be addressed by excluding (additional) confluent roots. Nonzero leading coefficients imply that higherorder formulae cannot be used for lowerorder problems right away. Still, lowerorder formulae can be recovered from higherorder ones by taking suitable limits (the world expert in this is Albert Rich). If one hits upon a factor like (m+n+1)*(m+n+2) in formula (8.1.2) of the paper, the obstacle can be moved out of the way by switching to a recurrence relation for the other exponent (see Rubi). Warning: Again, I haven't analyzed this issue formally to arbitray order, I simply find no reason to doubt this conclusion based on experience.
Martin.

